If the question is unclear, off topic or has other problems, please explain.

Consider the following sets that are infinitely countable and dense in $\mathbb{R}$

$T_1=\left\{\left.\frac{M_1(n_1)}{R_1(q_1)}\right|n_1,q_1\in\mathbb{Z}\right\}$ $,T_2=\left\{\left.\frac{M_2(n_2)}{R_2(q_2)}\right|n_2,q_2\in\mathbb{Z}\right\},...,$ $T_m=\left\{\left.\frac{M_m(n_m)}{R_m(q_m)}\right|n_m,q_m\in\mathbb{Z}\right\}$

Where $M_1(x),...,M_m(x)$ and $R_1(x),...,R_m(x)$ are unique functions. For example, $M_1(x)=2x+1$, $R_1(x)=2^x$, $M_2(x)=5x+1$, $R_2(x)=\ln(x)$ and so forth.

I want to construct a measure that compares the size of each set to the size of all the sets.

One user suggested

"It is indeed possible to get measures that behave in a nontrivial fashion on countable subsets, and indeed the axiom of choice or some weaker forms thereof are relevant here. Namely, one can get a finitely-additive (not $\sigma$-additive) measure $\xi$ on the set of subsets of $\mathbb{N}$ such that $\xi(S)=0$ for any finite set $S\subseteq\mathbb{N}$ and $\xi(S)=1$ for any cofinite subset (i.e., subset with a finite complement). Here $\xi$ takes only two values, zero or one. For sets that are neither finite nor cofinite, the measure behaves in a nontrivial way but always exactly one of a pair of complementary sets has measure one. Such measures are used in ultraproduct constructions."

I read articles on ultrafilters and ultraproducts, but the mathematics is far beyond my knowledge.

How does one assign measures, that use ultraproduct constructions, to sets $T_1,T_2,..,T_m$?

  • $\begingroup$ Why are my questions downvoted? $\endgroup$
    – Arbuja
    Jan 9, 2017 at 13:47
  • $\begingroup$ Why is my question off topic? What other sites can I ask this? $\endgroup$
    – Arbuja
    Jan 9, 2017 at 15:46
  • $\begingroup$ Any suggestions $\endgroup$
    – Arbuja
    Jan 10, 2017 at 15:18
  • $\begingroup$ Your post is just a bit difficult to read. The problem is that it is a bit long and it is easy to miss the question. $\endgroup$ Jan 14, 2017 at 0:24
  • $\begingroup$ @JacobMaibach I made edits. Let me know if its difficult to read. $\endgroup$
    – Arbuja
    Jan 14, 2017 at 0:37

1 Answer 1


Firstly, in the formal sense of the word measure, if you can measure all singleton sets $\{x\}$, then the measure of a countable set $T$ is $$\mu(T)=\sum_{x\in T}\mu(\{x\})$$ because measures are defined to be countably additive. That is, you assign a weight to every value, and just sum them up. That's not particularly interesting, especially when one realizes that, if one assigns a positive weight to more than countably many points, then you will end up with countable sets of infinite measure.

However, it seems like you are talking about the situation where you relax countable additivity to finite additivity. In this case, it is true that you can construct "measures" $\mu$ which act non-trivially on countable sets, but the issue is that ultraproduct constructions will not yield any meaningful information because, in such constructions, you are offered infinitely many choices and the measures of your sets will be determined by these choices.

Basically, the way the ultraproduct works is you try to create a set of "big" sets $U$ such that if a set contains a big set, it is itself big and such that, for any subset $X$ either $X$ or its complement is big, but not both. The way you do this is you see that, if you have some set of "big" sets $U$ satisfying the first condition that a set containing a big set is big and that there is no set $X$ such that both $X$ and $\mathbb R\setminus X$ are big. Either $U$ already contains every set or its complete, or there is some set $Y$ such that neither $Y$ nor $\mathbb R\setminus Y$ are yet declared big. You can check that arbitrarily declaring one of these and all sets containing it to be big maintains the conditions we demanded of $U$. Then, Zorn's lemma (which is equivalent to the axiom of choice) tells us, more or less, that we can do this process until we have chosen every set or its complement to be big, leaving us with a set $U$ with the desired properties. You would then declare $\mu(S)=1$ for any big $S$ and $\mu(S)=0$ otherwise.

The issue is that, when you're doing this, unless some of your $T_i$ contain each other, you can just choose whether they're big or small immediately, without relation to anything! So the result of this construction is completely arbitrary. That is to say, you could get any answer you wanted out of this.

  • $\begingroup$ Is there hope of using any other constructions $\endgroup$
    – Arbuja
    Jan 14, 2017 at 2:07
  • 1
    $\begingroup$ @Arbuja It depends what you're trying to measure; I can't really think of any good ideas for comparing countable dense sets to one another though - for countable nowhere dense sets, you could use asymptotic density (or otherwise measure the growth of the corresponding counting function), but countable dense is a bit weird (esp. since for two countable dense sets $T_1$ and $T_2$, there exists a continuous bijection $f:\mathbb R\rightarrow R$ with $f(T_1)=T_2$, and one may even choose $f$ to make $|f(x)-x|$ small, so it's hard to say such sets could be intrinsically different) $\endgroup$ Jan 14, 2017 at 2:36
  • $\begingroup$ My construction as shown in math.stackexchange.com/questions/2084047/… is similar to asymptotic density but the sets must be fully simplified in the numerator and denominator. Is there a chance of using my construction as a measure. $\endgroup$
    – Arbuja
    Jan 15, 2017 at 0:32
  • 1
    $\begingroup$ @Arbuja If you're committed to measuring sets, then no, that definitely won't work; your construction depends critically on the parameterization and there's really no way to fix it given how little structure countable dense sets have. Your construction really takes two sets $M$ and $R$ and asks how quickly the interval $(a,b)$ fills with numbers of the form $m/r$ for $m\in M$ and $r\in R\cap (-t,t)$ as we increase $t$. This is well-defined, as you can study the growth of the counting function in $t$. It measures some joint property of $M$ and $R$ as sets, but it's hard to say what it means. $\endgroup$ Jan 15, 2017 at 3:07
  • 1
    $\begingroup$ It seems like the idea is basically to, for every pair $M,R$ to define a function $f_{a,b}(t)=|\{m/r : m\in M, r\in R\cap [-t,t], m/r\in [a,b]\}$, then to compare them to each other. So, you're counting how many elements land in an interval as $t$ increases. Then you see whether that's faster or slower than the other sets. You could also use big O notation to more independently express the growth of $f$, which would at least distinguish sets with $0$ or $1$ as measure. (The notion of simplification is troublesome though - you can reduce sets of integers, but not so much sets of reals) $\endgroup$ Jan 15, 2017 at 16:19

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